A weak acid `HA` has `K_(a) = 10^(-6)`. What would be the molar ratio of this acid and its salt with strong base so that `pH` of the buffer solution is `5`?

Options

A. `1//10`
B. `10`
C. `1`
D. `2`

Explanation

To solve the problem of finding the molar ratio of a weak acid \( HA \) and its salt with a strong base so that the pH of the buffer solution is 5, we can follow these steps:

### Step 1: Understand the relationship between pH, pKa, and the concentrations of the acid and its salt.
The Henderson-Hasselbalch equation is given by:
\[ \text{pH} = \text{pKa} + \log\left(\frac{[A^-]}{[HA]}\right) \]
where:
- \( [A^-] \) is the concentration of the salt (the conjugate base),
- \( [HA] \) is the concentration of the weak acid.

### Step 2: Calculate pKa from Ka.
Given that \( K_a = 10^{-6} \), we can calculate \( pK_a \) using the formula:
\[ pK_a = -\log(K_a) \]
Substituting the value:
\[ pK_a = -\log(10^{-6}) = 6 \]

### Step 3: Substitute the known values into the Henderson-Hasselbalch equation.
We know the pH is 5, and we have calculated \( pK_a \) to be 6. Now we can substitute these values into the equation:
\[ 5 = 6 + \log\left(\frac{[A^-]}{[HA]}\right) \]

### Step 4: Rearrange the equation to solve for the ratio.
Rearranging the equation gives:
\[ \log\left(\frac{[A^-]}{[HA]}\right) = 5 - 6 = -1 \]

### Step 5: Convert from logarithmic form to ratio form.
To find the ratio \( \frac{[A^-]}{[HA]} \), we take the antilog:
\[ \frac{[A^-]}{[HA]} = 10^{-1} = 0.1 \]

### Step 6: Find the molar ratio of \( HA \) to \( A^- \).
The molar ratio of the weak acid \( HA \) to its salt \( A^- \) is the inverse of the ratio we just found:
\[ \frac{[HA]}{[A^-]} = \frac{1}{0.1} = 10 \]

### Final Answer:
The molar ratio of the weak acid \( HA \) to its salt \( A^- \) is \( 10:1 \).

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